Question

If \( a_1, a_2, a_3, \dots \) are in A.P., then the value of \( \begin{vmatrix} a_1 & a_2 & 1 \\ a_2 & a_3 & 1 \\ a_3 & a_4 & 1 \end{vmatrix} \) is equal to:

• A. \( a_4 - a_1 \)
• B. \( \frac{a_1 + a_4}{2} \)
• C. \( 1 \)
• D. \( \frac{a_2 + a_3}{2} \)
• E. \( 0 \)

Hint:

Determinants involving A.P. terms often resolve to zero because the "constant increase" creates linear dependency between rows.

Answer 1

Answer: (E)

To solve the given problem, we need to calculate the determinant of the matrix: 

\( a_1 \)	\( a_2 \)	1
\( a_2 \)	\( a_3 \)	1
\( a_3 \)	\( a_4 \)	1

Given that \( a_1, a_2, a_3, \dots \) are in Arithmetic Progression (A.P.), we have:

• \( a_2 = a_1 + d \)
• \( a_3 = a_1 + 2d \)
• \( a_4 = a_1 + 3d \)

Let's calculate the determinant:

\[\begin{vmatrix} a_1 & a_2 & 1 \\ a_2 & a_3 & 1 \\ a_3 & a_4 & 1 \end{vmatrix}\]

Using the determinant formula for a \(3 \times 3\) matrix, we have:

\[= a_1(a_3 - a_4) - a_2(a_2 - a_4) + 1(a_2a_4 - a_3a_3)\]

Substituting the expressions of \(a_2, a_3, a_4\) in terms of \(a_1\) and \(d\):

\[= a_1((a_1 + 2d) - (a_1 + 3d)) - (a_1 + d)((a_1 + d) - (a_1 + 3d)) + 1((a_1 + d)(a_1 + 3d) - (a_1 + 2d)^2)\]

Simplifying each term:

• \( a_3 - a_4 = (a_1 + 2d) - (a_1 + 3d) = -d \)
• \( a_2 - a_4 = (a_1 + d) - (a_1 + 3d) = -2d \)
• \( a_2a_4 = (a_1 + d)(a_1 + 3d) = a_1^2 + 4a_1d + 3d^2 \)
• \( a_3^2 = (a_1 + 2d)^2 = a_1^2 + 4a_1d + 4d^2 \)

Then, substitute and simplify:

\[= a_1(-d) - (a_1 + d)(-2d) + (a_1^2 + 4a_1d + 3d^2 - a_1^2 - 4a_1d - 4d^2)\]\[= -a_1d + 2(a_1d + d^2) + 0\]\[= -a_1d + 2a_1d + 2d^2\]\[= a_1d + 2d^2 - a_1d - 2d^2 = 0\]

Thus, the determinant is indeed \(0\). Therefore, the answer is:

\(0\).